# If $\int_0^\infty {e^{-\lambda t}f(t){\rm d}t} = 0$ for all $\lambda >0$ then $f=0$ a.e.?

I am studying a paper in which the author uses something like that:

Let $f$ be a bounded and Lebesgue measurable function. If $$\int_0^\infty {e^{-\lambda t}f(t)\,{\rm d}t} = 0 \qquad\text{for all} \qquad \lambda \gt0$$ then $f=0$ almost everywhere.

Could you point me to a proof of this theorem? It is very important for me.

Thanks!

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Thanks for help, i hope it works without continuity of $f$, becouse in my problem $f$ depends on transition function $P^t$ of right Markov process. – dawid Jun 25 '11 at 2:01
As Eric pointed out, this is related to the inverse Laplace transform. I didn't want to induce him to delete his answer, as this certainly is a valuable pointer. Off the top of my head, I don't remember the precise conditions for uniqueness, but this pointer should help looking for results. – t.b. Jun 25 '11 at 2:04

This is pretty easy to prove with Dynkin's multiplicative system theorem:

Theorem. Suppose $H$ is a vector space of bounded measurable functions on some measurable space $X$, which contains the constants and is closed under bounded pointwise convergence of sequences (i.e. if $f_n \in H$, $f_n \to f$ pointwise, and $|f_n| \le C$ for all $n$, then $f \in H$). Suppose $M \subset H$ is closed under pointwise multiplication, and let $\mathcal{G}$ be the $\sigma$-algebra generated by $M$ (i.e. the smallest $\sigma$-algebra on $X$ that makes all functions from $M$ measurable). Then $H$ contains all bounded $\mathcal{G}$-measurable functions.

The statement looks complicated, but the proof is elementary, and can be found, for instance, in Chapter II.8 of these notes, along with other references.

Now to prove your theorem, take $X = (0,\infty)$ with its Borel $\sigma$-algebra. We note it suffices to show $g(t) := e^{-t} f(t) = 0$ a.e. ($g$ is integrable, which is more convenient.)

Let $H$ be the set of all bounded measurable functions $h$ such that $\int_0^\infty h(t) g(t) dt = 0$. $H$ is clearly a vector space which contains the constants, and is closed under bounded convergence thanks to the dominated convergence theorem.

Let $M$ be the set of all functions of the form $h(t) = e^{-\lambda t}$, where $\lambda > 0$. By assumption, $M \subset H$, and $M$ is clearly closed under multiplication.

Finally, consider the $\sigma$-algebra $\mathcal{G}$ generated by $M$. Note that for any $h \in M$ and any open interval $(a,b) \subset \mathbb{R}$, we have by definition that $h^{-1}((a,b)) \in \mathcal{G}$. Taking $h(t) = e^{-t}$ and $(a,b) = (e^{-d}, e^{-c})$ we see that $(c,d) \in \mathcal{G}$. Since $\mathcal{G}$ contains all open intervals it must be the Borel $\sigma$-algebra of $(0, \infty)$.

So by Dynkin's theorem, $H$ contains all bounded Borel-measurable functions on $(0, \infty)$. In particular it contains ${g}$! This means that $\int_0^\infty g(t)^2 dt = 0$, so clearly $g = 0$ a.e.

The multiplicative system theorem is really a functional version of the better known $\pi$-$\lambda$ theorem, and acts as sort of a measurable-function analogue of the Stone-Weierstrass theorem. I've used it here as sort of an advertisement, since I think the theorem deserves to be more widely known given how useful it can be.

I'll also remark that the proof works just as well if we only assume that $\int_0^\infty e^{-\lambda t} f(t) dt = 0$ for large values of $\lambda$. Integer values of $\lambda$ would also suffice.

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thanks Nate, i think it is exactly what i need.. the uniqueness of the Laplace transformation seems to be naturally, but i noticed that the Learch's Theorem need the continuity, and your solution works without this assuption! – dawid Jun 25 '11 at 2:58
One question: Can we guarantee that $g$ is Borel measurable when $f$ is Lebesgue measurable? – Jose27 Jun 25 '11 at 3:55
@Jose27: If $f$ is Lebesgue measurable, then $f$ is a.e. equal to a Borel measurable function. (Proof: For $f$ an indicator function, this is the definition of the Lebesgue $\sigma$-algebra, so it also holds for simple functions. Now write $f$ as a limit of a sequence of simple functions.) So in fact we might as well assume $f$ is Borel measurable from the start. – Nate Eldredge Jun 25 '11 at 12:15

It looks like the result you're looking for is called Lerch's Theorem - essentially, if two functions have the same Laplace transform (or other linear integral transform) then their difference is a null function; i.e., its integral vanishes. In your case, one of the two functions is simply the zero function; since its Laplace transform is (trivially) identically zero, then any function whose Laplace transform is also identically zero must differ from it by a null function - or in other words, must be a null function itself; it is then (by definition!) zero a.e.

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Sorry if I ask: what exactly are the hypotheses of Lerch's theorem? The information in that link is completely useless: "functions", what is that supposed to mean? The only instances I found by googling require continuity of $f$ and thus don't answer the present question. – t.b. Jun 25 '11 at 2:37